Prove $(n^5-n)$ is divisible by 5 by induction. So I started with a base case $n = 1$. This yields $5|0$, which is true since zero is divisible by any non zero number. I let $n = k >= 1$ and let $5|A = (k^5-k)$. Now I want to show $5|B = [(k+1)^5-(k+1)]$ is true....
After that I get lost. 
I was given a supplement that provides a similar example, but that confuses me as well. 
Here it is if anyone wants to take a look at it:
Prove that for all n elements of N, $27|(10n + 18n - 1)$.
Proof: 
We use the method of mathematical induction. For $n = 1$, $10^1+18*1-1 = 27$. 
Since $27|27$, the statement is correct in this case.
Let $n = k = 1$ and let $27|A = 10k + 18k - 1$.
We wish to show that $27|B = 10k+1 + 18(k + 1) - 1 = 10k+1 + 18k + 17$.
Consider $C = B - 10A$ ***I don't understand why A is multiplied by 10.
$= (10k+1 + 18k + 17) - (10k+1 + 180k - 10)$  
$= -162k + 27
= 27(-6k + 1)$.
Then $27|C$, and $B = 10A+C$. Since $27|A$ (inductive hypothesis) and $27|C$, then
$B$ is the sum of two addends each divisible by $27$. By Theorem 1 (iii), $27|B$, and
the proof is complete.
 A: Hint $\ \ n^5\!-\!n  = n(n^2\!-\!1)(n^2\!-\!4\! +\! 5) = (n\!-\!2)(n\!-\!1)n(n\!+\!1)(n\!+\!2)+ 5n(n^2\!-\!1)$
Thus it suffices to show that $\,5\,$ divides a product of $\,5\,$ consecutive integers. In fact,  any sequence of $\,n\,$ consecutive naturals has an element divisible by $\,n\,$. This has a simple proof by induction:  shifting such a sequence  by one does not change its set of remainders mod $\,n,\,$ since it effectively replaces the old least element $\:\color{#C00}a\:$ by the new greatest element $\:\color{#C00}{a+n}$
$$\begin{array}{}& \color{#C00}a, &\!\!\!\! a+1, &\!\!\!\! a+2, &\!\!\!\! \cdots, &\!\!\!\! a+n-1  & \\
                  \to & &\!\!\!\! a+1,&\!\!\!\! a+2, &\!\!\!\! \cdots, &\!\!\!\! a+n-1, &\!\!\!\! \color{#C00}{a+n} \end{array}\qquad$$
Since $\: \color{#C00}{a\,\equiv\, a\!+\!n}\pmod n,\:$ the shift does not change the remainders in the sequence. Thus the remainders are the same as the base case $\ 0,1,2,\ldots,n-1\ =\: $ all $ $ possible remainders mod $\,n.\,$ Therefore the sequence has an element with remainder $\,0,\,$ i.e. an element divisible by $\,n.$
A: Your induction hypothesis is that $5\mid k^5-k$, which means that $k^5-k=5n$ for some integer $n$. Now
$$\begin{align*}
(k+1)^5-(k+1)&=\left(k^5+5k^4+10k^3+10k^2+5k+1\right)-(k+1)\\
&=k^5+5k^4+10k^3+10k^2+5k-k\\
&=(k^5-k)+5k^4+10k^3+10k^2+5k
\end{align*}$$
can you see why that must be a multiple of $5$?
A: Also note that
all the binomial coefficients
$C(p, k) = \frac{p!}{k!(p-k)!}$
when $p$ is a prime
are divisible by $p$
except for $k=0$ and $k=p$
(since $p$ is in the numerator
but not in the denominator).
This is enough to show by induction
that $n^p-n$ is divisible by
$p$ for all positive integers $n$:
True for $n=1$ and,
for the induction,
$\begin{align}
((n+1)^p-(n+1))-(n^p-n)
&=(n+1)^p-n^p-1\\
&= (n^p+(\text{terms divisible by p})+1)-n^p-1\\
&= (\text{terms divisible by p})
\end{align}$
.
A: Since $$(k+1)^5 - (k+1) = k^5 + 5k^4 + 10k^3 + 10k^2 + 5k - k$$, then $k^5 - k$ must be a multiple of $5$. And yes, it is by the fact that the unit digit of $k^5 - k$ will always be $0$. So the whole expression is really a multiple of $5$.
A: You can also use decomposition over polynomials :$\quad\displaystyle \Pi_k(n)=k!\binom nk=\prod\limits_{i=0}^{k-1} (n-i)$
$\displaystyle n^5-n = 30\times\left[ 4\binom n5+8\binom n4+5\binom n3+\binom n2\right]$
Since the binomial coefficients are integers, you get the divisibility by $30$ as a result.
This method can be generalised to other problems of the style, "show that $m$ divides $P(n)$".
A: Using modular arithmetic:
$$\begin{align}n&\equiv 0,1,2,3,4 \pmod 5 \\
n^5&\equiv 0,1,2,3,4 \pmod 5 \\
n^5-n &\equiv 0 \pmod 5.\end{align}$$
